Basis Of A Three Dimensional Space

Go back to  'Vectors and 3-D Geometry'

Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school

In the preceeding discussion, we talked about the basis of a plane. We can easily extend that discussion to observe that any three non-coplanar vectors can form a basis of three dimensional space:

In other words, any vector \(\vec r\) in 3-D space can be expressed as a linear combination of three arbitrary non-coplanar vectors. From this, it also follows that for three non-coplanar vectors \(\vec a,\vec b,\vec c,\) if their linear combination is zero, i.e, if

\[\lambda \vec a + \mu \vec b + \gamma \vec c = \vec 0\quad\qquad\qquad\left( {{\text{where}}\,\,\lambda ,\mu ,\gamma  \in \mathbb{R}} \right)\]

then \(\lambda ,\mu \,\,{\text{and}}\,\,\gamma \) must all be zero. To prove this, assume the contrary. Then, we have

\[\vec a = \left( { - \frac{\mu }{\lambda }} \right)\vec b + \left( { - \frac{\gamma }{\lambda }} \right)\vec c\]

which means that \(\vec a\) can be written as the linear combination of  \(\vec b\,\,{\text{and}}\,\,\vec c\). However, this would make \(\vec a,\,\,\vec b\,\,{\text{and}}\,\vec c\) coplanar, contradicting our initial supposition. Thus, \(\lambda ,\mu \,\,{\text{and}}\,\,\gamma \)  must be zero.

We finally come to what we mean by linearly independent and linearly dependent vectors.

Linearly independent vectors : A set of non-zero vectors \({\vec a_1},{\vec a_2},{\vec a_3}....,{\vec a_n}\)  is said to be linearly independent if

\[{\lambda _1}{\vec a_1} + {\lambda _2}{\vec a_2} + ... + {\lambda _n}{\vec a_n} = \vec 0\]

\[implies\,\,\,\,{\lambda _1} = {\lambda _2} = .... = {\lambda _n} = 0\]

Thus, a linear combination of linearly independent vectors cannot be zero unless all the scalars used to form the linear combination are zero.

Linearly dependent  vectors: A set of non-zero vectors  \({\vec a_1},{\vec a_2},{\vec a_3},....,{\vec a_n}\)  is said to be linearly dependent if there exist scalars \({\lambda _1},{\lambda _2}....{\lambda _n},\) not all zero such that,

\[{\lambda _1}{\vec a_1} + {\lambda _2}{\vec a_2} + ..... + {\lambda _n}{\vec a_n} = \vec 0\]

For example, based on our previous discussions, we see that

(i) Two non-zero, non-collinear vectors are linearly independent.

(ii) Two collinear vectors are linearly dependent

(iii) Three non-zero, non-coplanar vectors are linearly independent.

(iv) Three coplanar vectors are linearly dependent

(v) Any four vectors in 3-D space are linearly dependent.

You are urged to prove for yourself all these assertions.

Example – 5

Let \({\vec a},\vec b\,\,{\text{and}}\,\,\vec c\) be non-coplanar vectors. Are the vectors \(2\vec a - \vec b + 3\vec c,\,\,\vec a + \vec b - 2\vec c\,\,{\text{and}}\,\,\vec a + \vec b - 3\vec c\)  coplanar or non-coplanar?

Solution: Three vectors are coplanar if there exist scalars \(\lambda ,\mu  \in \mathbb{R}\) using which one vector can be expressed as the linear combination of the other two.

Let us try to find such scalars:

\[2\vec a - \vec b + 3\vec c = \lambda \left( {\vec a + \vec b - 2\vec c} \right) + \mu \left( {\vec a + \vec b - 3\vec c} \right)\]

\[ \Rightarrow  \quad  \left( {2 - \lambda  - \mu } \right)\vec a + \left( { - 1 - \lambda  - \mu } \right)\vec b + \left( {3 + 2\lambda  + 3\mu } \right)\vec c = \vec 0\]

Since \(\vec a,\vec b,\vec c,\) are non-coplanar, we must have

\[2 - \lambda  - \mu  = 0\]

\[ - 1 - \lambda  - \mu  = 0\]

\[3 + 2\lambda  + 3\mu  = 0\]

This system, as can be easily verified , does not have a solution for  \(\lambda \,\,{\text{and}}\,\,\mu \).

Thus, we cannot find scalars for which one vector can be expressed as the linear combination of the other two, implying the three vectors must be non-coplanar.

As an additional exercise, show that for three non-coplanar vectors \(\vec a,\,\,\vec b\,\,{\text{and}}\,\,\vec c\) , the vectors  \(\vec a - \,2\vec b\, + 3\vec c,\,\,\vec a - 3\vec b + 5\vec c\,\,\,and\,\,\, - 2\vec a - \,3\vec b\, - 4\vec c\)  are coplanar.

What is basis and dimension?

  • An important result in linear algebra is the following: Every basis for V has the same number of vectors. The number of vectors in a basis for V is called the dimension of V, denoted by dim(V).

What is a basis of a vector?

  • A vector basis of a vector space is defined as a subset of vectors in that are linearly independent and span

What is the dimensions of a set of vectors?

  • Dimension of a Vector Space If V is spanned by a finite set, then V is said to be finite-dimensional, and the dimension of V, written as dim V, is the number of vectors in a basis for V. The dimension of the zero vector space {0} is defined to be 0.

How to find the angle between two vectors in 3d?

  • a. Calculate the magnitude of the vectors.
  • b. Divide each vector by its vector magnitude to compute its unit vector.
  • c. Compute the dot product of the unit vectors.
  • d. Take the arcosine of the dot product of the unit vectors to get the angle between the vectors in radians.

What is the angle between two vectors?

  • The angle between two vectors, deferred by a single point, called the shortest angle at which you have to turn around one of the vectors to the position of co-directional with another vector.
Download SOLVED Practice Questions of Basis Of A Three Dimensional Space for FREE
Vectors
grade 11 | Questions Set 1
Vectors
grade 11 | Answers Set 1
Vectors
grade 11 | Questions Set 2
Vectors
grade 11 | Answers Set 2
Three Dimensional Geometry
grade 11 | Questions Set 2
Three Dimensional Geometry
grade 11 | Answers Set 2
Three Dimensional Geometry
grade 11 | Questions Set 1
Three Dimensional Geometry
grade 11 | Answers Set 1
Download SOLVED Practice Questions of Basis Of A Three Dimensional Space for FREE
Vectors
grade 11 | Questions Set 1
Vectors
grade 11 | Answers Set 1
Vectors
grade 11 | Questions Set 2
Vectors
grade 11 | Answers Set 2
Three Dimensional Geometry
grade 11 | Questions Set 2
Three Dimensional Geometry
grade 11 | Answers Set 2
Three Dimensional Geometry
grade 11 | Questions Set 1
Three Dimensional Geometry
grade 11 | Answers Set 1
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school